how to generate barcode in ssrs report For resonance, Im Yin is zero, so that in Software

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For resonance, Im Yin is zero, so that
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Fig. 12-43
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FREQUENCY RESPONSE, FILTERS, AND RESONANCE
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Clearly, as R ! 0, this expression reduces to that given for the pure LC tank. Substituting the numerical values produces a value of 0.791 for the radical; hence, !a 408:2 0:791 322:9 rad=s or fa 51:4 Hz
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12.17 Measurements on a practical inductor at 10 MHz give L 8:0 mH and Qind 40. (a) Find the ideal capacitance C for parallel resonance at 10 MHz and calculate the corresponding bandwidth . (b) Repeat if a practical capacitor, with a dissipation factor D Q 1 0:005 at 10 MHz, is cap used instead of an ideal capacitance.
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(a) From Section 12.14, 1 1 !a p q LC 1 Q 2 ind or C !2 L 1 a 1 Q 2 ind   31:6 pF 1 2 10 106 2 8:0 10 6 1 1600 1
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Using Section 12.15 to convert the series RL branch of Fig. 12-25 to parallel at the resonant frequency, ! L Rp R 1 Q2 a 1 Q2 ind ind Qind Then, from Section 12.13, ! !2 L ! Q 2 10 106 40 rad=s a a a ind 2 Rp 1 1600 Qa 1 Qind or 0.25 MHz. (b) The circuit is shown in Fig. 12-44; part (a) gives the resistance of the practical inductor as R !a L 4  Qind
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Also, from the given dissipation factor, it is known that 1 0:005 !a CRC The input admittance is     1 1 1 R !L j!C Yin j !C 2 RC R j!L RC R2 !L 2 R !L 2 which di ers from the input admittance for part (a) only in the real part. Since the imaginary part involves the same L and the same R, and must vanish at the same frequency, C must be the same as in part (a); namely, C 31:6 pF. For xed C, bandwidth is inversely proportional to resistance. With the practical capacitor, the net parallel resistance is R0 Rp RC Rp RC
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Fig. 12-44
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FREQUENCY RESPONSE, FILTERS, AND RESONANCE
[CHAP. 12
where Rp is as calculated in part (a).
Therefore,
Rp Rp ! L=Qind 1 Q2 ind 0 1 1 a 0:25 MHz R 1=!a C 0:005 RC 1 1 Q2 0:005 ind Qind 1 Q 2 ind 1 1600 0:005   1:2 1 1 40 1 1600
and so 0:30 MHz. A lossy capacitor has the same e ect as any loading resistor placed across the tank; the Qa is reduced and the bandwidth increased, while fa is unchanged.
12.18 A lossy capacitor, in the series-circuit model, consists of R 25  and C 20 pF. Obtain the equivalent parallel model at 50 kHz.
From Section 12.15, or by letting L ! 0 in Problem 12.6(a), Qs For this large Qs -value, Rp % Rs Q2 1010 M s Cp % Cs 20 pF 1 1 6370 !Cs Rs 2 50 103 20 10 12 25
12.19 A variable-frequency source of V 100 08 V is applied to a series RL circuit having R 20  and L 10 mH. Compute I for ! 0, 500, 1000, 2000, 5000 rad/s. Plot all currents on the same phasor diagram and note the locus of the currents.
Z R jXL R j!L Table 12-2 exhibits the required computations. With the phasor voltage at the angle zero, the locus of I as ! varies is the semicircle shown in Fig. 12-45. Since I VY, with constant V, Fig. 12-45 is essentially the same as Fig. 12-28(c), the admittance locus diagram for the series RL circuit. Table 12-2 !, rad/s 0 500 1000 2000 5000 XL ;  0 5 10 20 50 R;  20 20 20 20 20 Z;  20 08 20:6 14:048 22:4 26:578 28:3 458 53:9 68:208 5 08 4:85 14:048 4:46 26:578 3:54 458 1:86 68:208 I; A
12.20 The circuit shown in Fig. 12-46 is in resonance for two values of C when the frequency of the driving voltage is 5000 rad/s. Find these two values of C and construct the admittance locus diagram which illustrates this fact.
At the given frequency, XL 3 . Then the admittance of this xed branch is 1 0:147 j0:088 S 5 j3
Y1
CHAP. 12]
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